Let $A\subset \mathbb R^n$ and $M$ be the convex hull of the set $A$, e.g., $M:=Conv(A)$. The Minkowski function on $M$ is defined as follows \begin{align*} &f: \mathbb R^n \to \mathbb R\\ &f(x):=\inf\{t\ge 0: x\in tM\}. \end{align*} I was wondering if this function has a gradient respect to $x$ or not?
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$\begingroup$ If $M$ is the unit ball, then $f(x)=\lvert x\rvert$. So the short answer is no. $\endgroup$– Giuseppe NegroCommented Dec 11, 2015 at 11:01
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1$\begingroup$ $|x|$ has a gradient everywhere but $0$. More generally, $f$ is convex and therefore differentiable almost everywhere. $\endgroup$– Deane YangCommented Dec 11, 2015 at 12:53
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$\begingroup$ Here you can suppose that $A$ as an union of the ball. $\endgroup$– MSSHDCommented Dec 11, 2015 at 13:52
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$\begingroup$ Just an idea. Usually (I don't know the precise assumptions), the functional $f$ is a norm having $M$ as unit ball. Differentiability of $f$ is therefore related to the smoothness of the boundary of $M$. (I learned about this relationship here, look for "Gâteaux derivative"). Also, since $f$ is homogeneous of degree $1$, it is never going to be differentiable at the origin. $\endgroup$– Giuseppe NegroCommented Dec 11, 2015 at 16:36
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$\begingroup$ If $A$ is the union of a finite number of balls, then the boundary is $C^1$. In fact, I believe that it is $C^{1,1}$ and in fact smooth except on a lower dimensional set. $\endgroup$– Deane YangCommented Dec 11, 2015 at 18:37
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